Question

A capacitor of capacitance $4\mu \text{F}$ can withstand a maximum voltage of $10\text{V}$. Another capacitor of capacitance $3\mu \text{F}$ can withstand a maximum voltage of $8\text{V}$. If the capacitors are connected in parallel, the combination can withstand a maximum voltage of

• A. $10\text{V}$
• B. $8\text{V}$
• C. $4\text{V}$
• D. $5\text{V}$

Answer 1

The correct option is B $8\text{V}$

When connected in parallel to a battery of voltage $V$, both the capacitors will have the same potential difference across them, which is $V$.

If,
$\left(1\right).V>10\text{Volt}\Rightarrow$ Both capacitors will break-down.

$\left(2\right).10\text{Volt}>V>8\text{Volt}\Rightarrow$ Second capacitor will break- down.

$\left(3\right).V<8\text{Volt}\Rightarrow$ Both capacitors will work fine.

This simply means that for the combination of these two capacitors to not break-down, applied voltage must be upto $8\text{Volt}$.

Hence, option $\left(b\right)$ is correct.

Key concept- Breakdown voltage of capacitors with different capacity in their parallel connection.